2025 AMC 10A Problems/Problem 17: Difference between revisions

Revision as of 20:14, 6 November 2025

Problem

Let $N$ be the unique positive integer such that dividing $273436$ by $N$ leaves a remainder of $16$ and dividing $272760$ by $N$ leaves a remainder of $15$ . What is the tens digit of $N$ ?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution 1

The problem statement implies $N|273420$ and $N|272745.$ We want to find $N > 16$ that satisfies both of these conditions. Hence, we can just find the greatest common divisor of the two numbers. $\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45$ by the Euclidean Algorithm, so the answer is $\boxed{\text{(E) }4}.$

~Tacos_are_yummy_1

Solution 2

We have: 273436 ≡ 16 (mod N) 272760 ≡ 15 (mod N)

First we substract (273436 − 272760) = 676 ≡ 1 (mod N)

So N divides 675. Since N is greater than 16, possible divisors are all greater than 16 and are 25, 27, 45, 75, 135, 225, 675. We then check which ones work. If 273436 ≡ 16 (mod N), then 273420 must be divisible by N. 273420 ÷ 45 = 6076, so N = 45 works. So N = 45, and the tens digit is $\boxed{\text{(E) }4}.$ .

~Continuous_Pi

Solution 3

We get that: $273436 ≡ 16 (mod N)$ (Error compiling LaTeX. Unknown error_msg) and $272760 ≡ 15 (mod N)$ (Error compiling LaTeX. Unknown error_msg).

So we also have that: $273420 ≡ 0 (mod N)$ (Error compiling LaTeX. Unknown error_msg) and $272745 ≡ 0 (mod N).$ (Error compiling LaTeX. Unknown error_msg)

Notice that these are a multiple of $5$ . Now, we subtract these numbers to get $675$ . We see that $675 = 25 \dot 27.$ There is a factor of $5$ and $9$ . $273,420$ and $272745$ are multiples of $9$ as well, so our answer is just $\boxed{\text{(E) }45}$ .

~Aarav22 ~reformatting by Alzwang

Video Solution (In 2 Mins)

https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 26: / Line 26: @@
 We get that:
-≡ 16 (mod N) and
+<imath>273436 ≡ 16 (mod N)</imath> and
-≡ 15 (mod N).
+<imath>272760 ≡ 15 (mod N)</imath>.
 So we also have that:
-≡ 0 (mod N) and
+<imath>273420 ≡ 0 (mod N)</imath> and
-≡ 0 (mod N).
+<imath>272745 ≡ 0 (mod N).</imath>
+Notice that these are a multiple of <imath>5</imath>. Now, we subtract these numbers to get <imath>675</imath>.
+We see that <imath>675 = 25 \dot 27.</imath> There is a factor of <imath>5</imath> and <imath>9</imath>.
+<imath>273,420</imath> and <imath>272745</imath> are multiples of <imath>9</imath> as well, so our answer is just <imath>\boxed{\text{(E) }45}</imath>.
-Notice that these are a multiple of 5. Now, we subtract these numbers to get 675.
-We see that 675 = 25 * 27. There is a factor of 5 and 9.
-,420 and 272745 are multiples of 9 as well, so our answer is just <imath>\boxed{\text{(E) }4}5</imath>.
 ~Aarav22
+~reformatting by Alzwang
 == Video Solution (In 2 Mins) ==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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